In an AP, if Sn = 3n^2 + 5n and ak = 164, then find the value of k.
Answers
Solution:-
Given: Sn = 3n² + 5n
S₁ = 3(1)² + 5(1)
= 3 + 5
T₁ = 8
S₂ = 3(2)² + 5(2)
= 12 + 10
= 22
So, T₁ + T₂ = 22
8 + T₂ = 22
T₂ = 22 - 8
T₂ = 14
Now,
a = 8 and d = 14 - 8 = 6
As ak = 164
or a + (k-1)d = 164
8 + (k-1)6 = 164
6k - 6 = 164 - 8
6k = 164 + 6 - 8
6k = 162
k = 162/6
k = 27
Answer.
Given : Sn = 3n^2 + 5n.
Substitute n = 1:
⇒ S1 = 3(1)^2 + 5(1)
= 3 + 5
= 8.
Therefore, First term a1 = 8.
Substitute n = 2:
⇒ S2 = 3(2)^2 + 5(2)
= 12 + 10
= 22.
Therefore, Second term a2 = 22 - 8 = 14.
Now, We have to find the common difference.
Common difference d = a2 - a1
= 14 - 8
= 6.
Here, First term a = 8 and common difference d = 6.
Given that kth term of an AP is 164.
⇒ ak = a + (k - 1) * d
⇒ 164 = 8 + (k - 1) * 6
⇒ 164 = 8 + 6k - 6
⇒ 164 = 6k + 2
⇒ 162 = 6k
⇒ k = 162/6
⇒ k = 27.
Therefore, the value of k = 27.
Hope this helps!